Question

Question #52580

Answer 1

`"200 g MgCl"_2`

Explanation:

The first thing that you need to do here is to figure out how many grams of magnesium chloride would be produced by the reaction at `100%` yield.

To do that, use the molar mass of magnesium chloride

`3 color(red)(cancel(color(black)("moles MgCl"_2))) * "95.211 g"/(1color(red)(cancel(color(black)("mole MgCl"_2)))) = "285.63 g"`

This means that at `100%`yierld, you would expect the reaction to produce `"285.63 g"` of magnesium chloride.

However, you know that the reaction has a `73%` yield, which means that for every `"100 g"` of magnesium chloride that the reaction could theoretically produce, you only get `"73 g"`.

You can thus say that your reaction will produce

`285.63 color(red)(cancel(color(black)("g MgCl"_2color(white)(.)"in theory"))) * overbrace(("73 g MgCl"_2 color(white)(.)"produced")/(100color(red)(cancel(color(black)("g MgCl"_2color(white)(.)"in theory")))))^(color(blue)("= 73% yield"))`

`= color(darkgreen)(ul(color(black)("200 g MgCl"_2color(white)(.)"produced")))`

The answer is rounded to one significant figure, the number of sig figs you have for the number of moles of magnesium chloride that would be produced at `100%` yield.